Question

If ${\log _b}a.{\log _c}a + {\log _a}b.{\log _c}b + {\log _a}c.{\log _b}c = 3$ (where a, b, c are different positive real numbers ≠ 1) then find the value of $abc$ .

Answer 1

Hint : A logarithm, of a base b, is the power to which the base needs to be raised to yield a given number. We know that ${\log _b}a = \dfrac{{\log a}}{{\log b}}$ , so first convert the given logarithmic terms with bases to this form using this conversion. And then solve the remaining solution referring to the below mentioned formula.
Formulas used:
${\log _b}a = \dfrac{{\log a}}{{\log b}}$
If $x + y + z = 0$ , then ${x^3} + {y^3} + {z^3} = 3xyz$ and vice-versa.

Complete step-by-step answer :
We are given a logarithmic equation ${\log _b}a.{\log _c}a + {\log _a}b.{\log _c}b + {\log _a}c.{\log _b}c = 3$ where a, b, c are different positive real numbers ≠ 1.
We have to find the value of $abc$ .
As we already know that ${\log _b}a = \dfrac{{\log a}}{{\log b}}$ .
Therefore, Using the above conversion we are converting the logarithmic terms present in the given equation into this fractional form.
${\log _c}a = \dfrac{{\log a}}{{\log c}}$
${\log _a}b = \dfrac{{\log b}}{{\log a}}$
${\log _c}b = \dfrac{{\log b}}{{\log c}}$
${\log _a}c = \dfrac{{\log c}}{{\log a}}$
${\log _b}c = \dfrac{{\log c}}{{\log b}}$
On substituting all the obtained fractional terms in ${\log _b}a.{\log _c}a + {\log _a}b.{\log _c}b + {\log _a}c.{\log _b}c = 3$ , we get
$\left( {\dfrac{{\log a}}{{\log b}} \times \dfrac{{\log a}}{{\log c}}} \right) + \left( {\dfrac{{\log b}}{{\log a}} \times \dfrac{{\log b}}{{\log c}}} \right) + \left( {\dfrac{{\log c}}{{\log a}} \times \dfrac{{\log c}}{{\log b}}} \right) = 3$
$\Rightarrow \dfrac{{{{\left( {\log a} \right)}^2}}}{{\log b.\log c}} + \dfrac{{{{\left( {\log b} \right)}^2}}}{{\log a.\log c}} + \dfrac{{{{\left( {\log c} \right)}^2}}}{{\log a.\log b}} = 3$
Take out the LCM and convert the above left hand side into a single fraction, the LCM is $\log a.\log b.\log c$
$\Rightarrow \dfrac{{\log a \times {{\left( {\log a} \right)}^2} + \log b \times {{\left( {\log b} \right)}^2} + \log c \times {{\left( {\log c} \right)}^2}}}{{\log a.\log b.\log c}} = 3$
$\Rightarrow \dfrac{{{{\left( {\log a} \right)}^3} + {{\left( {\log b} \right)}^3} + {{\left( {\log c} \right)}^3}}}{{\log a.\log b.\log c}} = 3$
On cross multiplication, we get
$\Rightarrow {\left( {\log a} \right)^3} + {\left( {\log b} \right)^3} + {\left( {\log c} \right)^3} = 3 \times \left( {\log a.\log b.\log c} \right) = 3\log a.\log b.\log c$
As we can see, the above equation is in the form of ${x^3} + {y^3} + {z^3} = 3xyz$ , where x is $\log a$ , y is $\log b$ and z is $\log c$
Therefore, $x + y + z$ must be equal to zero which means $\log a + \log b + \log c = 0$
We know that $\log a + \log b$ is equal to $\log ab$
Therefore, $\log a + \log b + \log c = \log abc$
$\log abc = 0$
Sending the logarithm to the right hand side (as $\log abc$ is a common logarithm it will have a base 10)
$abc = {10^0} = 1$ (Anything to the power zero is equal to 1)
Therefore, the value of $abc$ is 1.
So, the correct answer is “1”.

Note : We know that ${\log _b}a = \dfrac{{\log a}}{{\log b}}$ , which can also be written as $\dfrac{1}{{\left( {\dfrac{{\log b}}{{\log a}}} \right)}} = \dfrac{1}{{{{\log }_a}b}}$ . And while finding the value of ${\log _b}a$ , confirm that b is always greater than zero and never equal 1; a must be a positive real number. If ${\log _b}a = k$ , then $a = {b^k}$